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Consider free, real scalar field theory in $d=1+3$ dimensions: $H = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi + \frac{1}{2} m^2 \phi^2$. The Hilbert space of this theory is known; it is just the vacuum plus each of the $n$-particle Hilbert spaces spanned by plane waves (something like $\mathbb{C} \times L^2(\mathbb{R}^3) \times L^2(\mathbb{R}^{3*2}) \times \cdots L^2(\mathbb{R}^{3n}) \times \cdots$)

Question: If we introduce a $\phi^4$ interaction term, is the Hilbert space known? If not, is at least known whether or not the free theory's Hilbert space is a subset of the interacting theory's Hilbert space?

(Motivation for this question: for a single particle in $3$ spatial dimensions, the Hilbert space of $H = \frac{P^2}{2m} + V$ is $L^2(\mathbb{R}^3)$ for a large class of potentials $V$, including the free case $V=0$.)

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I'm not an expert in this area... but most Hilbert spaces physicists use are separable. A large number of them are infinite-dimensional (exceptions e.g. spin systems). We know that all infinite-dimensional separable Hilbert spaces are isomorphic to one other. Hence there is only really 'one' Hilbert space. So the Hilbert space of the $\phi^4$ theory is the same as that of the free field theory. This is what allows us to do perturbation theory anyway - one can always write the states of the $\phi^4$ theory by using the basis states of the free field theory.. – nervxxx Apr 12 '13 at 3:16
Well, it'd be a Fock space (not a Hilbert space). That said, I don't believe it exists, since most of the time one just makes perturbative approximations when solving the quantum EOM. – Alex Nelson Apr 12 '13 at 4:11
@nervxxx It's true that Hilbert spaces of the same dimension are isomorphic. But these isomorphisms don't respect the domains where the operators are defined. (See the answer to this question:… ) This is why perturbation theory is tricky: the interacting states are complicated limits rather than simple sums. – user1504 Apr 12 '13 at 14:32
@AlexNelson: Fock spaces are Hilbert spaces. – user1504 Apr 12 '13 at 15:59

There are a few different ways of answering this question.

First, I should point out that there's a mathematical machine which takes a Euclidean path integral and spits out a Hilbert space with an algebra of field operators acting on it. So if you can write down the path integral for $\phi^4$ theory rigorously, you can turn a crank and get the Hilbert space. If this theory is interacting, then the Hilbert space and algebra of operators you get won't be isomorphic to the free field Hilbert space & field algebra. (The Hilbert spaces will be isomorphic -- although not canonically -- but the field operators will be defined on different domains, and the isomorphisms won't match these domains.)

The problem with this answer is that no one has constructed the path integral for 4d $\phi^4$ theory rigorously.

In fact, there are good reasons to suspect that the continuum $\phi^4$ theory in 4d doesn't exist. The theory appears to have a Landau pole; this means that if you start with some cutoff theory and try to remove the cutoff while holding the long distance correlation functions fixed, you discover that the $\phi^4$ coupling blows up to infinity at a finite cutoff scale. Which means that there's no consistent way of turning on a $\phi^4$ interaction in pure 4d scalar field theory.

An alternate way of saying this: It looks like the $\phi^4$ interaction is marginally irrelevant; $\phi^4$ theory looks just like free scalar field theory at long distances. The interactions go away. What this means in practice for the Hilbert spaces is that they're the same.

This doesn't mean that pure $\phi^4$ theory is useless, by the way. It can still make sense as an effective field theory approximation to some short distance theory. All it means is that the short distance theory can't be pure $\phi^4$ theory; there must be some new degrees of freedom in the UV completion.

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